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If $\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=-\hat{i}+2 \hat{k}$ then $|\overrightarrow{\mathrm{c}}|$. $\vec{a}$ is equal to :
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Verified Answer
The correct answer is:
$2 \sqrt{5} \hat{i}-2 \sqrt{5} \hat{j}+\sqrt{5} \hat{k}$
If $\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}$ and $\vec{c}=-\hat{i}+2 \hat{k}$
$$
\begin{aligned}
& |\overrightarrow{\mathrm{c}}|=\sqrt{(-1)^2+2^2}=\sqrt{1+4}=\sqrt{5} \\
& |\vec{c}| \cdot \vec{a}=\sqrt{5} \cdot(2 \hat{i}-2 \hat{j}+\hat{k}) \\
& \therefore|\vec{c}| \cdot \vec{a}=2 \sqrt{5} \hat{i}-2 \sqrt{5} \hat{j}+\sqrt{5} \hat{k}
\end{aligned}
$$
$$
\begin{aligned}
& |\overrightarrow{\mathrm{c}}|=\sqrt{(-1)^2+2^2}=\sqrt{1+4}=\sqrt{5} \\
& |\vec{c}| \cdot \vec{a}=\sqrt{5} \cdot(2 \hat{i}-2 \hat{j}+\hat{k}) \\
& \therefore|\vec{c}| \cdot \vec{a}=2 \sqrt{5} \hat{i}-2 \sqrt{5} \hat{j}+\sqrt{5} \hat{k}
\end{aligned}
$$
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